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Chapter 1: Q.10.10P (page 22)

${鈭�}_{n = 1}^{鈭�} \frac{{(- 1)}^{n} x^{2 n}}{{(2 n)}^{\frac{3}{2}}}$

Short Answer

Expert verified

The interval of convergence is $[- 1, 1]$

Step by step solution

01

Given Information

The power series is ${鈭�}_{n = 1}^{鈭�} \frac{{(- 1)}^{n} x^{2 n}}{{(2 n)}^{\frac{3}{2}}}$

02

Definition of the interval of convergence.

The Interval of convergence is the interval in which the power series is convergent.

03

Find the interval

The power series is ${鈭�}_{n = 1}^{鈭�} \frac{{(- 1)}^{n} x^{2 n}}{{(2 n)}^{\frac{3}{2}}}$

Let $蚁_{n} = |\frac{a_{n + 1}}{a_{n}}|$ ,

Substitute the value of the power series in the formula above, the equation becomes as follows,

$蚁_{n} = |\frac{a_{n + 1}}{a_{n}}|$

= \frac{\frac{x^{2 n + 2}}{{(2 n + 2)}^{\frac{3}{2}}}}{\frac{x^{2 n}}{{(2 n)}^{\frac{3}{2}}}}

$= |\frac{x^{2 n + 2} {(2 n)}^{\frac{3}{2}}}{x^{2 n} {(2 n + 2)}^{\frac{1}{2}}}|$

$= |\frac{x^{2} 2^{\frac{3}{2}}}{{(2 + \frac{2}{n})}^{\frac{3}{2}}}|$

Apply limits in the above equation,

$蚁 = \lim_{n 鈫� 鈭�} |\frac{x^{2} 2^{\frac{3}{2}}}{{(2 + \frac{2}{n})}^{\frac{3}{2}}}|$

= |x^{2}|

The power series is convergent for $蚁 < 1$ ,

Hence the given power series is convergent for $|x^{2}| < 1$ and divergent for $|x^{2}| > 1$ .

Now check the ends points,

When $x = 1$ series is ${鈭�}_{n = 1}^{鈭�} \frac{{(- 1)}^{n}}{n^{\frac{3}{2}}}$ , which is a convergent alternating series.

When $x = - 1$ series is ${鈭�}_{n = 1}^{鈭�} \frac{{(- 1)}^{n}}{n^{\frac{3}{2}}}$ , which is a convergent alternating series.

Hence the interval of convergence is $[- 1, 1]$ .

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Most popular questions from this chapter

${鈭�}_{n = 1}^{鈭�} 鈥� {(\sqrt{x^{2} + 1})}^{n} \frac{2^{n}}{3^{n} + n^{3}}$

Use power series to evaluate the function at the given point. Compare with computer results, using the computer to find the series, and also to do the problem without series. Resolve any disagreement in results (see Example 1). $e^{sinx} - \frac{1}{x^{3}} In (1 + x^{3} e^{x}) at x = 0.00035$ .

$x^{2} l n (1 - x)$

$\frac{s i n (\sqrt{x})}{\sqrt{x}}, x > 0$

$f (x) = 鈭� x, a = 25 .$

See all solutions

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